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3.160 \(\int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx\)

Optimal. Leaf size=211 \[ -\frac {\text {Li}_2\left (-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{4 \sqrt {a} \sqrt {b} d^2}+\frac {\text {Li}_2\left (-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{4 \sqrt {a} \sqrt {b} d^2}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d} \]

[Out]

-1/2*I*x*ln(1+(a-b)*exp(2*I*(d*x+c))/(a^(1/2)-b^(1/2))^2)/d/a^(1/2)/b^(1/2)+1/2*I*x*ln(1+(a-b)*exp(2*I*(d*x+c)
)/(a^(1/2)+b^(1/2))^2)/d/a^(1/2)/b^(1/2)-1/4*polylog(2,-(a-b)*exp(2*I*(d*x+c))/(a^(1/2)-b^(1/2))^2)/d^2/a^(1/2
)/b^(1/2)+1/4*polylog(2,-(a-b)*exp(2*I*(d*x+c))/(a^(1/2)+b^(1/2))^2)/d^2/a^(1/2)/b^(1/2)

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Rubi [A]  time = 0.53, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4588, 3321, 2264, 2190, 2279, 2391} \[ -\frac {\text {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{4 \sqrt {a} \sqrt {b} d^2}+\frac {\text {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{4 \sqrt {a} \sqrt {b} d^2}-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sec[c + d*x]^2)/(a + b*Tan[c + d*x]^2),x]

[Out]

((-I/2)*x*Log[1 + ((a - b)*E^((2*I)*(c + d*x)))/(Sqrt[a] - Sqrt[b])^2])/(Sqrt[a]*Sqrt[b]*d) + ((I/2)*x*Log[1 +
 ((a - b)*E^((2*I)*(c + d*x)))/(Sqrt[a] + Sqrt[b])^2])/(Sqrt[a]*Sqrt[b]*d) - PolyLog[2, -(((a - b)*E^((2*I)*(c
 + d*x)))/(Sqrt[a] - Sqrt[b])^2)]/(4*Sqrt[a]*Sqrt[b]*d^2) + PolyLog[2, -(((a - b)*E^((2*I)*(c + d*x)))/(Sqrt[a
] + Sqrt[b])^2)]/(4*Sqrt[a]*Sqrt[b]*d^2)

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
 (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
 - Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
 Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
 e, n}, x] && EqQ[c*d, 1]

Rule 3321

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c
 + d*x)^m*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)))/(b + 2*a*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)) - b*E^(2*I*k*Pi)*E^(
2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 4588

Int[(((f_.) + (g_.)*(x_))^(m_.)*Sec[(d_.) + (e_.)*(x_)]^2)/((b_) + (c_.)*Tan[(d_.) + (e_.)*(x_)]^2), x_Symbol]
 :> Dist[2, Int[(f + g*x)^m/(b + c + (b - c)*Cos[2*d + 2*e*x]), x], x] /; FreeQ[{b, c, d, e, f, g}, x] && IGtQ
[m, 0]

Rubi steps

\begin {align*} \int \frac {x \sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx &=2 \int \frac {x}{a+b+(a-b) \cos (2 c+2 d x)} \, dx\\ &=4 \int \frac {e^{i (2 c+2 d x)} x}{a-b+2 (a+b) e^{i (2 c+2 d x)}+(a-b) e^{2 i (2 c+2 d x)}} \, dx\\ &=\frac {(2 (a-b)) \int \frac {e^{i (2 c+2 d x)} x}{-4 \sqrt {a} \sqrt {b}+2 (a+b)+2 (a-b) e^{i (2 c+2 d x)}} \, dx}{\sqrt {a} \sqrt {b}}-\frac {(2 (a-b)) \int \frac {e^{i (2 c+2 d x)} x}{4 \sqrt {a} \sqrt {b}+2 (a+b)+2 (a-b) e^{i (2 c+2 d x)}} \, dx}{\sqrt {a} \sqrt {b}}\\ &=-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i \int \log \left (1+\frac {2 (a-b) e^{i (2 c+2 d x)}}{-4 \sqrt {a} \sqrt {b}+2 (a+b)}\right ) \, dx}{2 \sqrt {a} \sqrt {b} d}-\frac {i \int \log \left (1+\frac {2 (a-b) e^{i (2 c+2 d x)}}{4 \sqrt {a} \sqrt {b}+2 (a+b)}\right ) \, dx}{2 \sqrt {a} \sqrt {b} d}\\ &=-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 (a-b) x}{-4 \sqrt {a} \sqrt {b}+2 (a+b)}\right )}{x} \, dx,x,e^{i (2 c+2 d x)}\right )}{4 \sqrt {a} \sqrt {b} d^2}-\frac {\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 (a-b) x}{4 \sqrt {a} \sqrt {b}+2 (a+b)}\right )}{x} \, dx,x,e^{i (2 c+2 d x)}\right )}{4 \sqrt {a} \sqrt {b} d^2}\\ &=-\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}+\frac {i x \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{2 \sqrt {a} \sqrt {b} d}-\frac {\text {Li}_2\left (-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}-\sqrt {b}\right )^2}\right )}{4 \sqrt {a} \sqrt {b} d^2}+\frac {\text {Li}_2\left (-\frac {(a-b) e^{2 i (c+d x)}}{\left (\sqrt {a}+\sqrt {b}\right )^2}\right )}{4 \sqrt {a} \sqrt {b} d^2}\\ \end {align*}

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Mathematica [B]  time = 6.44, size = 512, normalized size = 2.43 \[ \frac {x \left (-\sqrt {a} \text {Li}_2\left (\frac {\sqrt {b} (1-i \tan (c+d x))}{i \sqrt {-a}+\sqrt {b}}\right )-\sqrt {a} \text {Li}_2\left (\frac {\sqrt {b} (i \tan (c+d x)+1)}{i \sqrt {-a}+\sqrt {b}}\right )+\sqrt {a} \text {Li}_2\left (-\frac {\sqrt {b} (\tan (c+d x)-i)}{\sqrt {-a}+i \sqrt {b}}\right )+\sqrt {a} \text {Li}_2\left (\frac {\sqrt {b} (\tan (c+d x)+i)}{\sqrt {-a}+i \sqrt {b}}\right )+4 i \sqrt {-a} c \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )-\sqrt {a} \log (1+i \tan (c+d x)) \log \left (\frac {\sqrt {-a}-\sqrt {b} \tan (c+d x)}{\sqrt {-a}-i \sqrt {b}}\right )+\sqrt {a} \log (1-i \tan (c+d x)) \log \left (\frac {\sqrt {-a}-\sqrt {b} \tan (c+d x)}{\sqrt {-a}+i \sqrt {b}}\right )-\sqrt {a} \log (1-i \tan (c+d x)) \log \left (\frac {\sqrt {-a}+\sqrt {b} \tan (c+d x)}{\sqrt {-a}-i \sqrt {b}}\right )+\sqrt {a} \log (1+i \tan (c+d x)) \log \left (\frac {\sqrt {-a}+\sqrt {b} \tan (c+d x)}{\sqrt {-a}+i \sqrt {b}}\right )\right )}{2 \sqrt {-a^2} \sqrt {b} d (\log (1-i \tan (c+d x))-\log (1+i \tan (c+d x))+2 i c)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*Sec[c + d*x]^2)/(a + b*Tan[c + d*x]^2),x]

[Out]

(x*((4*I)*Sqrt[-a]*c*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]] - Sqrt[a]*Log[1 + I*Tan[c + d*x]]*Log[(Sqrt[-a] -
Sqrt[b]*Tan[c + d*x])/(Sqrt[-a] - I*Sqrt[b])] + Sqrt[a]*Log[1 - I*Tan[c + d*x]]*Log[(Sqrt[-a] - Sqrt[b]*Tan[c
+ d*x])/(Sqrt[-a] + I*Sqrt[b])] - Sqrt[a]*Log[1 - I*Tan[c + d*x]]*Log[(Sqrt[-a] + Sqrt[b]*Tan[c + d*x])/(Sqrt[
-a] - I*Sqrt[b])] + Sqrt[a]*Log[1 + I*Tan[c + d*x]]*Log[(Sqrt[-a] + Sqrt[b]*Tan[c + d*x])/(Sqrt[-a] + I*Sqrt[b
])] - Sqrt[a]*PolyLog[2, (Sqrt[b]*(1 - I*Tan[c + d*x]))/(I*Sqrt[-a] + Sqrt[b])] - Sqrt[a]*PolyLog[2, (Sqrt[b]*
(1 + I*Tan[c + d*x]))/(I*Sqrt[-a] + Sqrt[b])] + Sqrt[a]*PolyLog[2, -((Sqrt[b]*(-I + Tan[c + d*x]))/(Sqrt[-a] +
 I*Sqrt[b]))] + Sqrt[a]*PolyLog[2, (Sqrt[b]*(I + Tan[c + d*x]))/(Sqrt[-a] + I*Sqrt[b])]))/(2*Sqrt[-a^2]*Sqrt[b
]*d*((2*I)*c + Log[1 - I*Tan[c + d*x]] - Log[1 + I*Tan[c + d*x]]))

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fricas [B]  time = 1.30, size = 3344, normalized size = 15.85 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sec(d*x+c)^2/(a+b*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

1/16*(-4*I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c*log(2*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a +
b)/(a - b)) + 2*cos(d*x + c) + 2*I*sin(d*x + c)) + 4*I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c*log(2*sqrt(-(2*
(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) + 2*cos(d*x + c) - 2*I*sin(d*x + c)) + 4*I*(a - b)*sqr
t(a*b/(a^2 - 2*a*b + b^2))*c*log(2*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) - 2*cos(d*
x + c) + 2*I*sin(d*x + c)) - 4*I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c*log(2*sqrt(-(2*(a - b)*sqrt(a*b/(a^2
- 2*a*b + b^2)) + a + b)/(a - b)) - 2*cos(d*x + c) - 2*I*sin(d*x + c)) + 4*I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b
^2))*c*log(2*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) + 2*cos(d*x + c) + 2*I*sin(d*x +
c)) - 4*I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c*log(2*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)
/(a - b)) + 2*cos(d*x + c) - 2*I*sin(d*x + c)) - 4*I*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*c*log(2*sqrt((2*(a
- b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) - 2*cos(d*x + c) + 2*I*sin(d*x + c)) + 4*I*(a - b)*sqrt(a
*b/(a^2 - 2*a*b + b^2))*c*log(2*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) - 2*cos(d*x +
c) - 2*I*sin(d*x + c)) + 4*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*dilog(1/2*((2*(a + b)*cos(d*x + c) + (2*I*a +
 2*I*b)*sin(d*x + c) - 4*((a - b)*cos(d*x + c) - (-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqr
t(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) - 2*a + 2*b)/(a - b) + 1) + 4*(a - b)*sqrt(a*b/(
a^2 - 2*a*b + b^2))*dilog(-1/2*((2*(a + b)*cos(d*x + c) - (2*I*a + 2*I*b)*sin(d*x + c) - 4*((a - b)*cos(d*x +
c) + (-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))
+ a + b)/(a - b)) + 2*a - 2*b)/(a - b) + 1) + 4*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*dilog(1/2*((2*(a + b)*co
s(d*x + c) + (-2*I*a - 2*I*b)*sin(d*x + c) - 4*((a - b)*cos(d*x + c) - (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2
 - 2*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) - 2*a + 2*b)/(a - b) + 1) +
 4*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*dilog(-1/2*((2*(a + b)*cos(d*x + c) - (-2*I*a - 2*I*b)*sin(d*x + c) -
 4*((a - b)*cos(d*x + c) + (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/
(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) + 2*a - 2*b)/(a - b) + 1) - 4*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*dil
og(1/2*((2*(a + b)*cos(d*x + c) + (2*I*a + 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x + c) + (I*a - I*b)*sin(d*x
 + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) - 2*a +
2*b)/(a - b) + 1) - 4*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*dilog(-1/2*((2*(a + b)*cos(d*x + c) - (2*I*a + 2*I
*b)*sin(d*x + c) + 4*((a - b)*cos(d*x + c) - (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*
(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) + 2*a - 2*b)/(a - b) + 1) - 4*(a - b)*sqrt(a*b/(a^2 -
2*a*b + b^2))*dilog(1/2*((2*(a + b)*cos(d*x + c) + (-2*I*a - 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x + c) + (
-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b
)/(a - b)) - 2*a + 2*b)/(a - b) + 1) - 4*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))*dilog(-1/2*((2*(a + b)*cos(d*x
+ c) - (-2*I*a - 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x + c) - (-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*
a*b + b^2)))*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) + 2*a - 2*b)/(a - b) + 1) + 4*(I*
(a - b)*d*x + I*(a - b)*c)*sqrt(a*b/(a^2 - 2*a*b + b^2))*log(-1/2*((2*(a + b)*cos(d*x + c) + (2*I*a + 2*I*b)*s
in(d*x + c) - 4*((a - b)*cos(d*x + c) - (-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt(-(2*(a
- b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) - 2*a + 2*b)/(a - b)) + 4*(-I*(a - b)*d*x - I*(a - b)*c)*
sqrt(a*b/(a^2 - 2*a*b + b^2))*log(1/2*((2*(a + b)*cos(d*x + c) - (2*I*a + 2*I*b)*sin(d*x + c) - 4*((a - b)*cos
(d*x + c) + (-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b +
 b^2)) + a + b)/(a - b)) + 2*a - 2*b)/(a - b)) + 4*(-I*(a - b)*d*x - I*(a - b)*c)*sqrt(a*b/(a^2 - 2*a*b + b^2)
)*log(-1/2*((2*(a + b)*cos(d*x + c) + (-2*I*a - 2*I*b)*sin(d*x + c) - 4*((a - b)*cos(d*x + c) - (I*a - I*b)*si
n(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) -
2*a + 2*b)/(a - b)) + 4*(I*(a - b)*d*x + I*(a - b)*c)*sqrt(a*b/(a^2 - 2*a*b + b^2))*log(1/2*((2*(a + b)*cos(d*
x + c) - (-2*I*a - 2*I*b)*sin(d*x + c) - 4*((a - b)*cos(d*x + c) + (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2
*a*b + b^2)))*sqrt(-(2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) + a + b)/(a - b)) + 2*a - 2*b)/(a - b)) + 4*(-I*(
a - b)*d*x - I*(a - b)*c)*sqrt(a*b/(a^2 - 2*a*b + b^2))*log(-1/2*((2*(a + b)*cos(d*x + c) + (2*I*a + 2*I*b)*si
n(d*x + c) + 4*((a - b)*cos(d*x + c) + (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*(a - b
)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) - 2*a + 2*b)/(a - b)) + 4*(I*(a - b)*d*x + I*(a - b)*c)*sqrt
(a*b/(a^2 - 2*a*b + b^2))*log(1/2*((2*(a + b)*cos(d*x + c) - (2*I*a + 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x
 + c) - (I*a - I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2))
 - a - b)/(a - b)) + 2*a - 2*b)/(a - b)) + 4*(I*(a - b)*d*x + I*(a - b)*c)*sqrt(a*b/(a^2 - 2*a*b + b^2))*log(-
1/2*((2*(a + b)*cos(d*x + c) + (-2*I*a - 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x + c) + (-I*a + I*b)*sin(d*x
+ c))*sqrt(a*b/(a^2 - 2*a*b + b^2)))*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) - 2*a + 2
*b)/(a - b)) + 4*(-I*(a - b)*d*x - I*(a - b)*c)*sqrt(a*b/(a^2 - 2*a*b + b^2))*log(1/2*((2*(a + b)*cos(d*x + c)
 - (-2*I*a - 2*I*b)*sin(d*x + c) + 4*((a - b)*cos(d*x + c) - (-I*a + I*b)*sin(d*x + c))*sqrt(a*b/(a^2 - 2*a*b
+ b^2)))*sqrt((2*(a - b)*sqrt(a*b/(a^2 - 2*a*b + b^2)) - a - b)/(a - b)) + 2*a - 2*b)/(a - b)))/(a*b*d^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sec \left (d x + c\right )^{2}}{b \tan \left (d x + c\right )^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sec(d*x+c)^2/(a+b*tan(d*x+c)^2),x, algorithm="giac")

[Out]

integrate(x*sec(d*x + c)^2/(b*tan(d*x + c)^2 + a), x)

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maple [B]  time = 0.60, size = 1003, normalized size = 4.75 \[ -\frac {a c x}{d \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {b c x}{d \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) a x}{2 d \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {\polylog \left (2, \frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right )}{2 d^{2} \left (-2 \sqrt {a b}-a -b \right )}-\frac {\polylog \left (2, \frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{2 \sqrt {a b}-a -b}\right )}{4 d^{2} \sqrt {a b}}-\frac {a \,c^{2}}{2 d^{2} \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {b \,c^{2}}{2 d^{2} \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{2 \sqrt {a b}-a -b}\right ) c}{2 d^{2} \sqrt {a b}}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) c}{d^{2} \left (-2 \sqrt {a b}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{2 \sqrt {a b}-a -b}\right ) x}{2 d \sqrt {a b}}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) x}{d \left (-2 \sqrt {a b}-a -b \right )}-\frac {\polylog \left (2, \frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) b}{4 d^{2} \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {\polylog \left (2, \frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) a}{4 d^{2} \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {i c \arctanh \left (\frac {2 \left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}+2 a +2 b}{4 \sqrt {a b}}\right )}{d^{2} \sqrt {a b}}-\frac {c^{2}}{2 d^{2} \sqrt {a b}}-\frac {c^{2}}{d^{2} \left (-2 \sqrt {a b}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) a c}{2 d^{2} \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {a \,x^{2}}{2 \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {b \,x^{2}}{2 \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {c x}{d \sqrt {a b}}-\frac {2 c x}{d \left (-2 \sqrt {a b}-a -b \right )}-\frac {x^{2}}{2 \sqrt {a b}}-\frac {x^{2}}{-2 \sqrt {a b}-a -b}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) b c}{2 d^{2} \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )}-\frac {i \ln \left (1-\frac {\left (a -b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{-2 \sqrt {a b}-a -b}\right ) b x}{2 d \sqrt {a b}\, \left (-2 \sqrt {a b}-a -b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sec(d*x+c)^2/(a+b*tan(d*x+c)^2),x)

[Out]

-1/d/(a*b)^(1/2)/(-2*(a*b)^(1/2)-a-b)*a*c*x-1/d/(a*b)^(1/2)/(-2*(a*b)^(1/2)-a-b)*b*c*x-1/2*I/d^2/(a*b)^(1/2)/(
-2*(a*b)^(1/2)-a-b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/2)-a-b))*a*c-1/4/d^2/(a*b)^(1/2)/(-2*(a*b)^(1/2)-
a-b)*polylog(2,(a-b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/2)-a-b))*b-1/2/d^2/(a*b)^(1/2)/(-2*(a*b)^(1/2)-a-b)*a*c^2-1
/2/d^2/(a*b)^(1/2)/(-2*(a*b)^(1/2)-a-b)*b*c^2-1/4/d^2/(a*b)^(1/2)/(-2*(a*b)^(1/2)-a-b)*polylog(2,(a-b)*exp(2*I
*(d*x+c))/(-2*(a*b)^(1/2)-a-b))*a-I/d^2*c/(a*b)^(1/2)*arctanh(1/4*(2*(a-b)*exp(2*I*(d*x+c))+2*a+2*b)/(a*b)^(1/
2))-1/2*I/d^2/(a*b)^(1/2)*ln(1-(a-b)*exp(2*I*(d*x+c))/(2*(a*b)^(1/2)-a-b))*c-I/d^2/(-2*(a*b)^(1/2)-a-b)*ln(1-(
a-b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/2)-a-b))*c-1/2*I/d/(a*b)^(1/2)*ln(1-(a-b)*exp(2*I*(d*x+c))/(2*(a*b)^(1/2)-a
-b))*x-I/d/(-2*(a*b)^(1/2)-a-b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/2)-a-b))*x-1/2/d^2/(a*b)^(1/2)*c^2-1/
d^2/(-2*(a*b)^(1/2)-a-b)*c^2-1/2/d^2/(-2*(a*b)^(1/2)-a-b)*polylog(2,(a-b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/2)-a-b
))-1/4/d^2/(a*b)^(1/2)*polylog(2,(a-b)*exp(2*I*(d*x+c))/(2*(a*b)^(1/2)-a-b))-1/2/(a*b)^(1/2)/(-2*(a*b)^(1/2)-a
-b)*a*x^2-1/2/(a*b)^(1/2)/(-2*(a*b)^(1/2)-a-b)*b*x^2-1/d/(a*b)^(1/2)*c*x-2/d/(-2*(a*b)^(1/2)-a-b)*c*x-1/2*I/d/
(a*b)^(1/2)/(-2*(a*b)^(1/2)-a-b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/2)-a-b))*b*x-1/2*I/d^2/(a*b)^(1/2)/(
-2*(a*b)^(1/2)-a-b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/2)-a-b))*b*c-1/2*I/d/(a*b)^(1/2)/(-2*(a*b)^(1/2)-
a-b)*ln(1-(a-b)*exp(2*I*(d*x+c))/(-2*(a*b)^(1/2)-a-b))*a*x-1/2/(a*b)^(1/2)*x^2-1/(-2*(a*b)^(1/2)-a-b)*x^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sec \left (d x + c\right )^{2}}{b \tan \left (d x + c\right )^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sec(d*x+c)^2/(a+b*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate(x*sec(d*x + c)^2/(b*tan(d*x + c)^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x}{{\cos \left (c+d\,x\right )}^2\,\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(cos(c + d*x)^2*(a + b*tan(c + d*x)^2)),x)

[Out]

int(x/(cos(c + d*x)^2*(a + b*tan(c + d*x)^2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sec ^{2}{\left (c + d x \right )}}{a + b \tan ^{2}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sec(d*x+c)**2/(a+b*tan(d*x+c)**2),x)

[Out]

Integral(x*sec(c + d*x)**2/(a + b*tan(c + d*x)**2), x)

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